How do you find the radius of convergence #Sigma n^n/(n!) x^n# from #n=[1,oo)#?
1 Answer
Mar 3, 2017
Explanation:
Use the ratio test, which states that
#L=lim_(nrarroo)abs((n+1)^(n+1)/((n+1)!)x^(n+1)(n!)/n^n1/x^n)#
Simplifying:
#L=lim_(nrarroo)abs((n+1)^(n+1)/n^n((n!)/((n+1)n!))x^(n+1)/x^n)#
#L=lim_(nrarroo)abs((n+1)^(n+1)/(n+1)1/n^nx)#
Bringing the
#L=absxlim_(nrarroo)abs((n+1)^n/n^n)#
#L=absxlim_(nrarroo)abs(((n+1)/n)^n)#
This is a well-known limit that approaches
#L=eabsx#
The posted series will converge when
#eabsx<1#
#absx<1/e#
Thus the radius of convergence is